mathschallenge.net

Problem

Consider the following results.

10¹ + 1 = 11
10² 1 = 99 = 9 11
10³ + 1 = 1001 = 91 11
10⁴ 1 = 9999 = 909 11
10⁵ + 1 = 100001 = 9091 11

Prove that 10ⁿ1 is divisible by 11 if n is even and 10ⁿ+1 is divisible by 11 if n is odd.

Solution

As (x1)ⁿ = (x1)(xⁿ¹ + x^x2 + ... + 1), we get:

10^2k1 = 100^k1 = (1001)(100^k1 + ... + 1) = 99Q 0 mod 11.

That is, 10ⁿ1 is divisible by 11 if n is even.

As 10^2k 1 mod 11, we can multiply both sides of the congruence by 10 to obtain 10^2k+1 10 -1 mod 11; that is, 10^2k+1+1 0 mod 11.

Hence we prove that 10ⁿ+1 is divisible by 11 if n is odd.

Prove that 100ⁿ+1 is divisible by 101 iff n is odd, 1000ⁿ+1 is divisible by 1001 iff n is odd.
What about 10000ⁿ+1? Generalise.